**You are welcome to ask questions by sending e-mail to the author**

## Rayleigh criterion of instability

**Question: **

What one should reguire for instability of a parallel flow: vorticity maximum or velocity inflection (the latter includes vorticity minima)?

**Answer: **

Vortex picture of KH instability in Sect 2.1.1 suggests that one should require maximum of the vorticity modulus, minimum is stable. This is the subject of the Fjortoft refinement of the Rayleigh criterium: if U''(0)=U(0)=0 then a necessary (but not sufficient) condition for inviscid instability is that somewhere UU''<0.

**Question Date: **

Mon, 02/12/2018

## Problem 1.11 - free kick

**Question: **

The question seems to address passing the ball around the wall of defenders and so the horizontal deflection of interest seems to be different than the one calculated in the solution - it is the difference between the radius of the ball's trajectory and the distance of the straight trajectory from the center of that circle (the straight trajectory is a chord on that circle of some 1km radius).

**Answer: **

Answer: This is true but the difference is small because the curvature radius is much larger than all the other distances.

**Your Name: **

Guy Gaziv

**Question Date: **

Tue, 02/10/2015

## Problem 1.8

**Question:**

In the solution to the Problem 1.8, it is said that the air density must be small to consider the oscillations quasi-static. That statement follows by the adiabatic law. Adiabaticity presumes fast changes. Is there a contradiction?

**Answer: **

No, there is no contradiction, but extra explanation is needed indeed (will be added to the second edition). The period of oscillations is the radius of the bubble divided by the speed of sound in the water, as follows from (4.12). That period is supposed to be much larger than the radius of the bubble divided by the speed of sound in the air (which is assured by much lower air density), so that the pressure can be considered uniform inside the bubble. On the other hand, the period is supposed to be much smaller than the heat exchange time, which is the radius squared divided by the thermal diffusivity. In distinction from the viscosity (see Section 1.4.3), the diffusivity in a liquid requires the transfer of molecules and can be estimated as the speed of sound times the mean free path. Therefore, the heat exchange is negligible when the radius is much larger than the mean free path. That allows us to use the adiabatic law.

**Your Name:**

Anna Frishman

**Question Date: **

Tue, 12/02/2014

## Quasi-momentum and velocity potential

**Question:**

Is the relation between the quasi-momentum and the velocity potential on page 30 is valid for compressible flows?

**Answer: **

No, it is valid only for incompressible flows. More accurately, the density is supposed to be non-changing during the pressure pulse which brings the body into motion.

**Your Name: **

AF, UL

**Question Date: **

Thu, 12/04/2014

## Flow past a wing

**Question:**

Why it is claimed in sect 1.5.4 that fluid elements, separated by a wing, do not meet at the trailing edge? Why the upper one travels faster?

**Answer:**

Pressure difference *P _{2}-P_{1 }*between upper and lower sides of a wing is proportional to the wing thickness

*h*, or in other words, to the small parameter

*h/L*, where

*L*is the width. The same is true for the velocity difference since by the Bernoulli theorem

*P _{2}-P_{1 }*=ρ(v

_{1}

^{2}-v

_{2}

^{2})/2=ρ(v

_{1}-v

_{2})(v

_{1}+v

_{2})/2≈ρ(v

_{1}-v

_{2})u.

*P*≈ρu∫(v

_{2}-P_{1 })dx_{1}-v

_{2})dx.

Now, velocity difference is proportional to *h/L*, while the difference in the path lengths is proportional to *h ^{2}/L*. Therefore, for a slender wing, the upper fluid element reaches the trailing edge before the lower one. That can be see in the photograph in the Additional Material "

**Flow past a wing"**.

**Your Name:**

**Question Date:**

## Zero resistance force

**Question: **

In Sect 11, Landau & Lifshits relate the absense of frag force to the fact that velocity of the potential flow decays too fast at infinity so that no energy flux goes there. Do you agree?

**Answer: **

Answer. It is a little bit more complicated. First, it is not the flux of energy but that of quasi-momentum which determines the force. Second, the flow pattern by itself is not that relevant to the existence of the flux at infinity and force acting on the body. What is important is left-right symmetry of the potential flow and the breakdown of this symmetry by the wake, which provides for a nonzero drag. Of course, wake has a slower decay of velocity at large distances than potential dipole flow, so symmetry breaking and law of decay are related.

**Your Name: **

GK

**Question Date: **

Wed, 04/24/2013

## Convection threshold

**Question: **

Is there a simpler way to derive the convection criterium (1.9)?

**Answer:**

Answer. Yes. It is simply the statement that the gain in potential energy gdz must exceed the decrease in thermal energy cpdT.

**Your Name: **

during Berkeley colloquium

**Question Date: **

Mon, 04/01/2013

## Induced mass

**Question:**

Is the induced mass always smaller than the displaced mass (like half for a sphere)?

**Answer:**

Answer. Not at all. Induced mass can be much larger (for a thin disc moving perpendicular to its plane) and much smaller (for a needle moving "end on") than the displaced mass.

**Your Name: **

Weizmann student

**Question Date:**

Tue, 07/24/2012

## Kelvin theorem and angular momentum

**Question: **

Angular momentum is proportional to the velocity circulation for a rotating cylinder. Does it mean that the conservation of the velocity circulation, i.e. Kelvin theorem, is equivalent to the conservation of angular momentum (see Feynman Lectures)?

**Answer: **

With all due respect to Feynman, the answer is no, angular momentum has a local density while Kelving theorem is a nonlocal conservation law.

**Your Name: **

KR

**Question Date: **

Thu, 07/19/2012

## Hamiltonian in Lagrangian coordinates

**Question: **

If Lagrangian description of an ideal fluid is a canonical Hamiltonian formalism (sect 1.3.4), why it cannot be used for statistical theory? Introducing the distribution in the phase space etc.

**Answer: **

Answer. Using the canonical structure in Lagrangian variables is not going to be easy. It is quite an unusual formalism with a Hamiltonian that changes its form with time - indeed, the form of how pressure (and enthalpy) depends on the coordinates changes with time. For quasi-momentum in Sect 1.3 it did not matter since it exploits only spatial invariance. For other purposes, it is quite a headache. I am also not sure how to introduce an ensemble to make sense of a phase-space distribution. Which does not mean that using this formalism for statistics is impossible, just nobody did it before to the best of my knowledge.

**Your Name:**

KR

**Question Date: **

Thu, 05/03/2012

## Phase and group velocity of surface waves

**Question: **

Is the symbol

means the same in the figures on pp 94 and 145?

**Answer: **

Answer: No. On p.95, it is the wavelength that corresponds to the minimum of the phase velocity, which takes place exactly when the phase and group velocities are equal:

On p. 145, it is the wavelength that corresponds to the minimum of the group velocity which is the speed of the caustics.

I apologize for denoting different quantities by the same symbol.

**Your Name:**

Anonymous

**Question Date:**

Mon, 01/23/2012

## Pascal Law

**Question:**

Is not this law just a simple consequence of randomness and isotropy of molecular motion?

**Answer: **

Pascal Law in hydrostatics does not appeal to the molecular motion. In hydrostatics this is just the requirement of the absense of net momentum flux (i.e. mechanical equilibrium). Note that we do not assume isotropy; for instance, fluid can be under gravity and have its density height-dependent. The fact that we can use Pascal law not only in hydrostatics by also in motion is based on the separation of scales between molecular motion and fluid flows. For this law to be true, we need to assume that deviations from thermal equilibrium (which any inhomogeneous flow creates) are small.

Indeed, one can extend the proof for the case of motion as follows: Consider a prism inside the fluid. The mass inside is proportional to the area that is to the squared linear size. If the forces per unit area are different then they do not sum into zero and we obtain the resulting force proportional to the linear size of the prism cross-section. Now, if we consider the limit where the cross-section shrinks to zero we obtain infinite acceleration which is impossible. We then conclude that the pressure must be the same in all directions.A finite acceleration of a fluid element then appears proportional to the gradient of pressure, so that the resulting force on the prism is proportional to the squared linear size. This argument is similar to the proof that the stress tensor is symmetric in Sect. 1.4.2.

**Your Name:**

HM Sher

**Question Date: **

Tue, 11/01/2011

## Pitot tube

**Question: **

What if the velocity is not directed parallel to the tube and actually the B point is on some other streamline.

**Answer: **

Answer: Pitot tube measures the velocity on the streamline which comes to the point B.

**Your Name: **

HM Sher

**Question Date: **

Tue, 11/01/2011

## Flow in a curved space

**Question: **

How fluid flows in a curved space?

**Answer: **

Answer: Probably the main difference with the flat space is that Galilean invariance no longer works. See e.g. "Turbulence on hyperbolic plane..".

**Question Date: **

Mon, 10/03/2011

## Fluctuations of the wake boundary

**Question: **

On p. 67, you mention the fluctuations of the wake boundary. What is known about the statistics of these fluctuations?

**Answer:**

Answer: Not much, I'm afraid. There are some old experimental data, mentioned at the end of Sect 5.8 of Statistical Hydrodynamics I by Monin & Yaglom. No theory.

**Question Date:**

Sun, 10/02/2011

## Turbulence generation by shear

**Question:**

Formula (2.4) and Figure 2.5 show that shear dV0x/dz>0 increases the energy of the secondary flow v1 if v1xv1z<0. How turbulence can be sustained by shear if in a random flow streamlines with opposite signs of v1xv1z are equally probably, so on average there is no energy flux from shear to turbulence?

**Answer:**

Answer: Turbulence is not isotropic when the mean flow has a shear. It is clear that the mean value of <v1xv1z> must be negative since it is the mean vertical flux of horizontal momentum, which must flow from high- to low-V0 region. It means that in a shear-generated turbulence, more streamlines are oriented as shown in Figure 2.5 than at a right angle to it.

**Question Date: **

Fri, 08/19/2011

## Acoustic pulse in 2d

**Question: **

Section 2.3.1 describes acoustic pulses which propagate withour changing shape in 1d and 3d. What about 2d?

**Answer: **

Apparently, two-dimensional axially symmetric wave equation,

does not have a solution of the form

Indeed, one can consider a short pulse of a localized force, which adds δ(t)δ(r) to the rhs of the equation. By the treatment analogous to (4.41)-(4.42) one can obtain that h=0 for r>ct and

for r<ct. In other words, the shape changes. See Fluid Mechanics by Landau & Lifshits, Sect. 71 for a general solution.

Generally, any localized perturbation produces large time asymptotics of the form

**Your Name: **

KR

**Question Date: **

Sun, 07/17/2011

## About drag crisis

**Question:**

What is the explanation for the drag crisis?

**Answer:**

Answer: Drag crisis is a sudden drop of the drag coefficient from 0.5 to 0.15 when Re increases from 15000 to 25000. It was discovered by Eiffel (in 1912) and explained by Prandtl as due to an onset of turbulence inside the boundary layer. Turbulent boundary layer entrains more fluid from outside, has thus more momentum and separates later downstream, which decreases the wake area and the drag.

**Question Date:**

Thu, 02/03/2011

## About cumulative jet, Exercise 1.15

**Question: **

Detonation of explosives actually determines the compression speed of the cone, that is the velocity U normal to the cone in its reference frame. How to relate this speed to the speed of the cumulative jet?

**Answer: **

Answer: All jets have velocity V on their surface. In the reference frame moving with the velocity Vcosα the cone moves normal to itself with the speed Vtgα=U. In this frame the cumulative jet moves with the speed v=V(1+1/cosα)=U(1+cosα)/sinα. Apparently, the smaller the angle, the larger is the jet speed. In most shells U is of order 2 km/sec while v can be tens of km/sec.

**Question Date: **

Sun, 12/19/2010